Question

我一直在尝试编写一个将前缀表达式转换为Postfix表达式的代码。

到目前为止，这是我所做的（注意我是新的，因此可能效率不高！）

/***************************************************
 NOTE: This code fails to realise the precedence
    of parentheses and $ (exponential) expressions
****************************************************/

import java.util.Scanner;

class Stack{
    //Private Declarations
    private int top;
    private String a[] = new String [100];
    //Public Declarations
    public Stack(){
        top = -1;
        a[0] = "\0";
    }
    public void push(String x){
        if (top != 99){
            a[++top] = x;
        }
        else{
            System.out.println("Stack overflow");
        }
    };
    public String pop(){
        if (top == -1){
            System.out.println("\nStack empty!");
            return "\n";
        }
        else{
            return a[top--];
        }
    };
    public int ret_top(){
        return top;
    };
}


public class Prefix2Postfix {
    public static void main(String[] args) {
        //Declaration
        Scanner in = new Scanner (System.in);
        Stack op = new Stack ();
        Stack sym = new Stack ();
        char ch;
        int i;
        String exp, str, str1, str2, str3;

        //Taking input from the user
        System.out.println("Enter the prefix expression : ");
        exp = in.next();

        for (i=0; i<exp.length(); i++){
            ch = exp.charAt(i);
            if((ch == '+')
             ||(ch == '-')
             ||(ch == '*')
             ||(ch == '/')
             ||(ch == '$')){
                str = Character.toString(ch);
                op.push(str);
            }
            else{
                str = Character.toString(ch);
                sym.push(str);
            if (sym.ret_top() == 1){
                str1 = sym.pop();
                str2 = sym.pop();
                str3 = op.pop();
                str1 = str2.concat(str1);
                str1 = str1.concat(str3);
                sym.push(str1);
            }
            }
        }

        //Output
        str = sym.pop();
        System.out.println("After conversion to postfix" + ": " + str);
        in.close();
    }
}

正如我已经在代码中评论过的那样，我的代码在前缀表达式的情况下无法实现优先级。

例如：

Infix   : a + (b*c)
Prefix  : +a*bc 
Postfix : abc*+ //Expected output 
Output I get : ab*c+

我使用的逻辑有什么问题，或者我需要添加什么？您能否建议一些方法，我的代码可以正常工作？如果有人可以提供帮助，我将不胜感激。

注意：我们的教授建议我们编写自己的堆栈类和代码。另请注意，这是一项家庭作业。

提前致谢。

修改为了确保我的STACK正常工作，我编写了以下代码，它的工作完好无损。

import java.util.Scanner;

class STACK {
    //Private Declarations
    private int top;
    private int elem[] = new int [100];
    //Public Declarations
    public STACK (){
        top = -1;
        elem[0] = 0;
    }
    public void push(int x){
        if (top != 99){
            elem[++top] = x;
        }
        else{
            System.out.println("Stack overflow");
        }
    };
    public int pop(){
        if (top == -1){
            System.out.println("Stack empty!");
            return 0;
        }
        else{
            return elem[top--];
        }
    };
}

public class StackPushPop {
    public static void main(String[] args) {
        STACK s = new STACK();
        Scanner in = new Scanner (System.in);
        int choice, x;
        do{
            System.out.println("Menu Options :");
            System.out.println("1 -> Push an element");
            System.out.println("2 -> Pop an element");
            System.out.println("3 -> Empty complete stack");
            System.out.println("Any other input for exit");
            System.out.println("Your choice : ");
            choice = in.nextInt();
            switch(choice){
                case 1: 
                    System.out.println("\nEnter element : ");
                    x = in.nextInt();
                    s.push(x);
                    break;
                case 2:
                    System.out.print("\nPopping element : ");
                    x = s.pop();
                    if (x != 0){
                        System.out.println(x);
                    }
                    break;
                case 3:
                    System.out.println("\nEmptying stack!");
                    x = 1;
                    while (x!= 0){
                        x = s.pop();
                        if(x != 0){
                            System.out.print(x + "  ");
                        }
                    }
                    break;
                default:
                    choice = 0;
            }

        }while (choice != 0);
    }
}

修改我终于设法创建了一个完美无缺的程序。

import java.util.Scanner;

class STACK{
    private int top, MAX;
    private String a[] = new String [1000];
    public STACK(){
        top = -1;
        MAX = 1000;
        a[0] = "";
    }
    public void push(String x){
        if (top <= MAX-1){
            a[++top] = x;
        }
        else{
            System.out.println("Stack overflow");
        }
    };
    public String pop(){
        if (top == -1){
            System.out.println("\nStack empty!");
            return "\n";
        }
        else{
            return a[top--];
        }
    };
    public int getTop(){
        return top;
    };
}

public class Prefix2Postfix_STACK{
    static boolean isOperator (char ch){
        switch (ch){
            case '+':
            case '-':
            case '*':
            case '/':
            case '$':
                        return true;
            default :
                        return false;
        }
    }
    public static void main(String[] args) {
        //declarations
        Scanner in = new Scanner (System.in);
        String exp;
        int i;
        STACK s = new STACK ();
        String exp_str[] = new String[100];
        String postfix_exp = "\n";

        //input
        System.out.println("Enter prefix expression (No spaces or brackets) : ");
        exp = in.next();

        //create a string array of all characters but in reverse
        for(i=0; i<=exp.length()-1; i++){
            exp_str[exp.length()-1-i]=Character.toString(exp.charAt(i));
        }

        //computing postfix:
        i=0;
        do{ 
            if (!isOperator(exp_str[i].charAt(0)))
                s.push(exp_str[i]);
            else{
                String str1 = s.pop();
                String str2 = s.pop();
                str1 = str1 + str2 + exp_str[i];
                postfix_exp = str1;
                s.push(str1);
            }
            i++;
        }while(s.getTop()>=0 && i!=exp.length());

        //Output
        System.out.println("After converting to postfix : " + postfix_exp);
        in.close();
    }
}

Answer 1

您编写的代码只能将操作员从左向右移动。向正确的方向轻推：

插入条件以检查符号中的运算符前缀字符串并先解决它们！

这可能会解决您的问题。

将表达式从前缀转换为后缀（面临优先级问题）

1 个答案: